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If $L : C^2[a,b] \rightarrow C^0[a,b] $ is s.t. $L y(t) = \ddot y(t) +p \dot y + q y(t) $ and $L$ is invertible then $L^{-1}$ has at most countable eigenvalues and they accumulate in $0$.

Why countable? And why should they accumulate in $0$?

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  • $\begingroup$ As stated, I don't think L can be invertible, since for any $y(a), y'(a)$ there is a unique $y(t)$ with $Ly = 0$. $\endgroup$ – Robert Lewis Sep 27 '13 at 1:37

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